# Precalculus

## Analytic Geometry # Distance from a Point to a Line on a Plane

The distance from a point M(x0, y0) to a straight line Ax + By + C = 0 in the plane is determined by the formula

$d = \frac{\left|{Ax_0 + By_0 + C}\right|}{\sqrt{A^2 + B^2}}$

Here (x0, y0) is the coordinate of the point M and A, B, C are the coefficients of the equation of the straight line.

This formula calculates the perpendicular distance d = MN between the point and the line. The formula can be proved as follows.

## Proof

Let point N(x1, y1) be the foot of the perpendicular drawn from the point M(x0, y0) to the line Ax + By + C = 0. Find the slope of the line Ax + By + C = 0:

$Ax + By + C = 0, \Rightarrow By = - Ax - C, \Rightarrow y = -\frac{A}{B}x - \frac{C}{B}.$

Therefore, the slope is equal to $$k = -\frac{A}{B}.$$

Since the lines MN and Ax + By + C = 0 are perpendicular, the slope of line MN is equal to

$k_{MN} = -\frac{1}{k} = -\frac{1}{-\frac{A}{B}} = \frac{B}{A}.$

Hence the equation of the straight line MN has the form

$y = \frac{B}{A}x + B_{MN}.$

Find now the value of BMN. Since point M(x0, y0) belongs to this line, then its coordinates satisfy the equation:

$y_0 = \frac{B}{A}x_0 + B_{MN}, \Rightarrow B_{MN} = y_0 - \frac{B}{A}x_0.$

So

$y = \frac{B}{A}x + y_0 - \frac{B}{A}x_0.$

Point N(x1, y1) also belongs to the line MN, so its coordinates also satisfy this equation:

$y_1 = \frac{B}{A}x_1 + y_0 - \frac{B}{A}x_0.$

From here we get

$y_1 - y_0 = \frac{B}{A}\left({x_1 - x_0}\right), \Rightarrow \frac{y_1 - y_0}{B} = \frac{x_1 - x_0}{A}.$

We denote

$\frac{y_1 - y_0}{B} = \frac{x_1 - x_0}{A} = q.$

Then

$y_1 - y_0 = Bq,\;x_1 - x_0 = Aq.$

The distance between points M and N is determined by the formula

$d\left({M,N}\right) = \sqrt{\left({x_1 - x_0}\right)^2 + \left({y_1 - y_0}\right)^2 }.$

Therefore

$d\left({M,N}\right) = \sqrt{\left({Aq}\right)^2 + \left({Bq}\right)^2} = |q|\sqrt{A^2 + B^2}.$

Notice, that

$y_1 - y_0 = Bq,\; x_1 - x_0 = Aq, \;\Rightarrow y_1 = y_0 + Bq,\; x_1 = x_0 + Aq.$

Taking into account that the point N(x1, y1) lies on the line Ax + By + C = 0, we obtain

$Ax_1 + By_1 + C = 0, \Rightarrow A\left({x_0 + Aq}\right) + B\left({y_0 + Bq}\right) + C = 0.$

Find the value of q from this equation:

$Ax_0 + A^2q + By_0 + B^2q + C = 0, \Rightarrow q\left({A^2 + B^2}\right) = - \left({Ax_0 + By_0 + C}\right), \Rightarrow q = -\frac{Ax_0 + By_0 + C}{A^2 + B^2}.$

Substituting the value of $$q$$ into the formula for the distance, we get

$d\left({M,N}\right) = |-\frac{Ax_0 + By_0 + C}{A^2 + B^2}|\sqrt{A^2 + B^2}.$

Using some algebra we get the final expression:

$d\left({M,N}\right) = \frac{\left|{Ax_0 + By_0 + C}\right|}{\sqrt{A^2 + B^2}}$

## Solved Problems

### Example 1.

Find the distance from the origin to the straight line $${6x - 8y + 5 = 0}.$$

Solution.

The distance from a point M(x0, y0) to the line Ax + By + C = 0 is determined by the formula

$d = \frac{\left|{Ax_0 + By_0 + C}\right|}{\sqrt{A^2 + B^2}}.$

Substitute the coordinates of the origin O(0,0) and the coefficients of the line:

$d = \frac{\left|{6 \cdot 0 - 8 \cdot 0 + 5}\right|}{\sqrt{6^2 + \left({-8}\right)^2}} = \frac{5}{\sqrt{100}} = \frac{5}{10} = \frac{1}{2}.$

### Example 2.

Find the length of the altitude $$BD$$ in a triangle with vertices $${A\left({-3,0}\right)},$$ $${B\left({2,5}\right)},$$ and $${C\left({3,2}\right)}.$$

Solution.

Write the equation of the line $$AC$$ in two-point form:

$\frac{y - y_A}{y_C - y_A} = \frac{x - x_A}{x_C - x_A},\; \Rightarrow \frac{y - 0}{2 - 0} = \frac{x + 3}{3 + 3},\; \Rightarrow \frac{y}{2} = \frac{x + 3}{6},$

and convert it into the general form:

$6y = 2\left({x+3}\right),\;\Rightarrow 2x - 6y + 6 = 0,\; \Rightarrow x - 3y + 3 = 0.$

Calculate now the distance from point $$B$$ to line $$AC,$$ that is, the length of the segment $$BD:$$

$d = \frac{\left|{Ax_B + By_B + C}\right|}{\sqrt{A^2 + B^2}} = \frac{\left|{1\cdot2 - 3\cdot5 + 3}\right|}{\sqrt{1^2 + \left({-3}\right)^2}} = \frac{\left|{2-15+3}\right|}{\sqrt{10}} = \frac{\left|{-10}\right|}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}.$

### Example 3.

Find the distance between two parallel lines $$Ax + By + C_1 = 0$$ and $$Ax + By + C_2 = 0.$$

Solution.

Let the point $$M\left({x_0,y_0}\right)$$ belong to the line $$Ax + By + C_1 = 0.$$ The distance from point M to the straight line $$Ax + By + C_2 = 0$$ is given by

$d = \frac{\left|{Ax_0 + By_0 + C_2}\right|}{\sqrt{A^2 + B^2}}.$

Since the point M belongs to the first line, we have

$Ax_0 + By_0 + C_1 = 0, \Rightarrow Ax_0 + By_0 = -C_1.$

Substituting this into the previous expression, we get

$d = \frac{\left|{-C_1 + C_2}\right|}{\sqrt{A^2 + B^2}} = \frac{\left|{C_2 - C_1}\right|}{\sqrt{A^2 + B^2}}.$

### Example 4.

A line drawn through the origin at the same distance from points $$M\left({3,3}\right)$$ and $$N\left({5,1}\right).$$ Find the equation of this line.

Solution.

Since the line passes through the origin, its equation is $$Ax + By = 0.$$ The distances from points M and N to the line are equal, therefore

$d = \frac{\left|{Ax_M + By_M}\right|}{\sqrt{A^2 + B^2}} = \frac{\left|{Ax_N + By_N}\right|}{\sqrt{A^2 + B^2}},\Rightarrow \left|{3A + 3B}\right| = \left|{5A + B}\right|.$

Expanding the absolute values, we get two solutions (two straight lines):

$3A + 3B = \pm\left({5A + B}\right).$

Find the equation of the first line:

$3A + 3B = 5A + B, \Rightarrow 2A = 2B, \Rightarrow B = A, \Rightarrow x + y = 0.$

The equation of the second straight line is written as

$3A + 3B = -5A - B, \Rightarrow 8A = -4B, \Rightarrow B = -2A, \Rightarrow x - 2y = 0.$

The location of these lines is shown in the figure below.